\documentclass[a4paper]{article}
\usepackage[margin=1in]{geometry}
\usepackage{ctex}
\usepackage{tikz}
\usepackage{color}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{fontspec}
\usepackage{amsthm}
\usepackage{xltxtra}
\usepackage{pgfplots}
\usepackage{pgfplotstable}
\usepackage{mflogo}
\usepackage{texnames}
\usepackage{graphicx}
\usepackage{titlesec}

\setmainfont{Times New Roman}
\setCJKmainfont[BoldFont=SimHei,ItalicFont = SimSun]{SimSun}

\newtheorem{definition}{Definition}[section]%定义
\newtheorem{theorem}{Theorem}[section]%定理
\newtheorem{axiom}{Axiom}[section]%公理
\newtheorem{lemma}{Lemma}[section]%引理
\newtheorem{proposition}{Proposition}[section]%命题
\newtheorem{corollary}{Corollary}[section]%推论
\newtheorem{remark}{Remark}[section]%注

\title{\heiti\zihao{2} 习题17.3}
\author{中书君}
\date{\today}
\begin{document}
\maketitle
\section{利用 Green 公式计算下列积分}
\subsection{$\int\limits_{L}(x+y) \mathrm{d} x-(x-y) \mathrm{d} y$, 其中 $L$ 为椭圆 $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$ 的逆时针方向}
\textbf{解}\quad
$P=(x+y),Q=(y-x)$则
$$
	\begin{aligned}
		\int\limits_{L}(x+y) \mathrm{d} x-(x-y) \mathrm{d} y & =\iint\limits_S\left(\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y}\right)\mathrm{d}x\mathrm{d}y \\
		                                                     & =-2\iint\limits_S\mathrm{d}x\mathrm{d}y                                                                         \\
		                                                     & =-2\pi ab
	\end{aligned}
$$

\subsection{$\int\limits_{L}\left(\mathrm{e}^{x} \sin y-m y\right) \mathrm{d} x+\left(\mathrm{e}^{x} \cos y-m\right) \mathrm{d} y$, 其中 $L$ 为上半圆周 $x^{2}+y^{2}=a x$ 沿 $x$ 增加的方向}
\textbf{解}\quad
\begin{tikzpicture}
	\draw[->] (-0.2,0)--(3.2,0)node[right]{$x$};
	\draw[->] (0,-0.2)--(0,3.2)node[above]{$y$};
	\draw[thick,red,<-] (2,0) arc(0:180:1);
	\draw[blue,<-](0,0)--(2,0);
\end{tikzpicture}

如图连接圆弧的两点,$P=\mathrm{e}^x\sin y -my, Q=\mathrm{e}^x\cos y - m$.
$$
	\begin{aligned}
		\int\limits_{L}\left(\mathrm{e}^{x} \sin y-m y\right) \mathrm{d} x+\left(\mathrm{e}^{x} \cos y-m\right) \mathrm{d} y & =-\iint\limits_S\left(\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y}\right)\mathrm{d}x\mathrm{d}y \\
		                                                                                                                     & +\int_{a}^{0}\mathrm{e}^x\sin0-m\cdot 0\mathrm{d}x                                                               \\
		                                                                                                                     & =-\iint\limits_Sm\mathrm{d}x\mathrm{d}y                                                                          \\
		                                                                                                                     & =-\dfrac{ma^2\pi}{8}
	\end{aligned}
$$



\subsection{$\int\limits_{L}\left(5 x y-\mathrm{e}^{x} \sin y\right) \mathrm{d} y+\mathrm{e}^{x} \cos y \mathrm{~d} x$, 其中 $L: x=\sqrt{2 y-y^{2}}$ 方向沿 $y$ 增大方向}
\textbf{解}\quad
\begin{tikzpicture}
	\draw[->] (-0.2,0)--(3.2,0)node[right]{$x$};
	\draw[->] (0,-0.2)--(0,3.2)node[above]{$y$};
	\draw[red,thick,->](0,0)arc(-90:90:1);
	\draw[blue,->](0,2)--(0,0);
\end{tikzpicture}

$P=\mathrm{e}^x\cos y,Q=5xy-\mathrm{e}^x\sin y$

连接圆弧的两个端点
$$
	\begin{aligned}
		\int\limits_{L}\left(5 x y-\mathrm{e}^{x} \sin y\right) \mathrm{d} y+\mathrm{e}^{x} \cos y \mathrm{~d} x & =\iint\limits_S\left(\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y}\right)\mathrm{d}x\mathrm{d}y \\
		                                                                                                         & -\int_2^0-\sin y\mathrm{d}y                                                                                     \\
		                                                                                                         & =\iint\limits_S5y\mathrm{d}x\mathrm{d}y+\cos 2-1                                                                \\
		                                                                                                         & =5\int_{-\pi/2}^{\pi/2}\mathrm{d}\theta\int_0^1\rho+\rho^2\sin\theta\mathrm{d}\rho+\cos 2-1                     \\
		                                                                                                         & =\dfrac{5\pi}{2}+\cos 2-1
	\end{aligned}
$$

\subsection{$\int\limits_{L} \ln x \mathrm{~d} x+\ln y \mathrm{~d} y$, 其中 $L$ 为抛物线 $y=x^{2}$ 上从 $A(1,1)$ 到 $B(2,4)$ 的一段弧}
\textbf{解}\quad
\begin{tikzpicture}
	\draw[->] (-0.2,0)--(3.2,0)node[right]{$x$};
	\draw[->] (0,-0.2)--(0,3.2)node[above]{$y$};
	\draw[->,thick,red,domain=1:2]plot(\x, \x^2);
	\draw[->,thick,blue](2,4)--(1,1);
\end{tikzpicture}

$P=\ln x,Q=\ln y$
连接曲线的两个端点,令此直线上的$x=1+t,y=1+3t$.
$$
	\begin{aligned}
		\int\limits_{L} \ln x \mathrm{~d} x+\ln y \mathrm{~d} y & =\iint\limits_S\left(\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y}\right)\mathrm{d}x\mathrm{d}y-\int_1^0\ln (1+t)+3\ln(1+3t)\mathrm{d}t \\
		                                                        & =\iint\limits_S0\mathrm{d}x\mathrm{d}y - \int_1^0\ln (1+t)+3\ln(1+3t)\mathrm{d}t                                                                        \\
		                                                        & =10\ln 2-4
	\end{aligned}
$$

推荐使用做法:
\begin{tikzpicture}
	\draw[->] (-0.2,0)--(3.2,0)node[right]{$x$};
	\draw[->] (0,-0.2)--(0,3.2)node[above]{$y$};
	\draw[->,thick,red,domain=1:2]plot(\x, \x^2);
	\draw[->,thick,blue](2,4)--(1,4);
	\draw[->,thick,blue](1,4)--(1,1);
\end{tikzpicture}

\subsection{$\oint\limits_{L}\left(x^{2}-2 y\right) \mathrm{d} x+\left(3 x+y \mathrm{e}^{y}\right) \mathrm{d} y$, 其中 $L$ 为由直线 $y=0, x+2 y=2$ 及圆弧 $x^{2}+y^{2}=1$ 所围成的区域 $D$ 的边界，方向：从点 $(2,0)$ 开始沿 $x+2 y=2$ 到点 $(0,1)$, 然后再沿圆弧
$x^{2}+y^{2}=1$ 到点 $(-1,0)$,最后沿点 $x$ 轴回到点 $(2,0)$.}
\textbf{解}\quad
\begin{tikzpicture}
	\draw[->] (-0.2,0)--(3.2,0)node[right]{$x$};
	\draw[->] (0,-0.2)--(0,3.2)node[above]{$y$};
	\draw[thick,red,->](0,1)arc(90:180:1);
	\draw[thick,red,->](-1,0)--(2,0);
	\draw[thick,red,<-](0,1)--(2,0);
\end{tikzpicture}

$P=x^2-2y,Q=3x+y\mathrm{e}^y$
$$
	\begin{aligned}
		\oint\limits_{L}\left(x^{2}-2 y\right) \mathrm{d} x+\left(3 x+y \mathrm{e}^{y}\right) \mathrm{d} y & =\iint\limits_S\left(\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y}\right)\mathrm{d}x\mathrm{d}y \\
		                                                                                                   & =\iint\limits_S5\mathrm{d}x\mathrm{d}y                                                                          \\
		                                                                                                   & =\dfrac{5\pi}{4}+5
	\end{aligned}
$$


\section{计算积分 $\int\limits_{L} \dfrac{x \mathrm{~d} y-y \mathrm{~d} x}{a^{2} x^{2}+b^{2} y^{2}}, L$ 为任意不经过原点的光滑封闭曲线,逆时针方向}
\textbf{解}\quad
$P=\dfrac{-y}{a^2x^2+b^2y^2},Q=\dfrac{x}{a^2x^2+b^2y^2}$
若$L$为围成拓扑单连通的约当曲线,则有
$$
	\begin{aligned}
		\int\limits_{L} \dfrac{x \mathrm{~d} y-y \mathrm{~d} x}{a^{2} x^{2}+b^{2} y^{2}} & =\iint\limits_S\left(\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y}\right)\mathrm{d}x\mathrm{d}y \\
		                                                                                 & =\iint\limits_S0\mathrm{d}x\mathrm{d}y                                                                          \\
		                                                                                 & =0
	\end{aligned}
$$
若为拓扑复连通,则构造曲线$L1:a^2x^2+y^2b^2=c^2$,为右侧,方向向内.
$$
	\begin{aligned}
		\int\limits_{L} \dfrac{x \mathrm{~d} y-y \mathrm{~d} x}{a^{2} x^{2}+b^{2} y^{2}} & =0-\int\limits_{L1}\dfrac{x\mathrm{d}y-y\mathrm{d}x}{c^2} \\
		                                                                                 & =\iint\limits_S\dfrac{2}{c^2}\mathrm{d}x\mathrm{d}y       \\
		                                                                                 & =\dfrac{2\pi}{ab}
	\end{aligned}
$$
\section{用 Green 公式计算下列曲线围成的面积}
\subsection{双纽线 $\left(x^{2}+y^{2}\right)^{2}=a^{2}\left(x^{2}-y^{2}\right)$( 提示 : 令 $y=x \tan \theta $)}
\textbf{解}$1^{\circ}$\quad
令$y=x \tan \theta $得$x^2=a^2(\cos^2\theta-\sin^2\theta)\cos^2\theta$,则有
$$
	\left\{\begin{array}{l}
		x=a\sqrt{\cos 2\theta}\cos\theta \\
		y=a\sqrt{\cos 2\theta}\sin\theta
	\end{array}\right.\qquad\theta\in\left[-\dfrac{\pi}{4},\dfrac{\pi}{4}\right]
$$
令$\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y}=1$,则可令$Q=\dfrac{x}{2},P=-\dfrac{y}{2}$.取$y$轴右侧的部分:
$$
	\begin{aligned}
		S & =2\iint\limits_S\mathrm{d}x\mathrm{d}y                                                                           \\
		  & =2\iint\limits_S\left(\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y}\right)\mathrm{d}x\mathrm{d}y \\
		  & =\oint\limits_Lx\mathrm{d}y-y\mathrm{d}x                                                                         \\
		  & =\int_{-\pi/4}^{\pi/4}a^2\cos 2\theta\mathrm{d}\theta                                                            \\
		  & =a^2
	\end{aligned}
$$
P.S.链式法则并不容易得到结果.

\textbf{解}$2^{\circ}$\quad
令$y=\tan\theta x$,解得$x=\dfrac{a\sqrt{1-\tan^2\theta}}{1+\tan^2\theta}$
$$
	\begin{aligned}
		x\mathrm{d}y-y\mathrm{d}x & =x(\tan\theta)[x(\tan\theta)+\tan\theta x'(\tan\theta)]\mathrm{d}\tan\theta-\tan\theta x(\tan\theta)\cdot x'\tan\theta\mathrm{d}\tan\theta \\
		                          & =(x^2(\tan\theta))\mathrm{d}\tan\theta
	\end{aligned}
$$
从而
$$
	\begin{aligned}
		S & =4\cdot\dfrac{1}{2}\int_0^1a^2\dfrac{1-\tan^2\theta}{(1+\tan^2\theta)^2}\mathrm{d}\tan\theta=2a^2\int_0^{\pi/4}\cos 2\theta\mathrm{d}\theta \\
		  & =a^2
	\end{aligned}
$$

\subsection{笛卡尔叶形线 $x^{3}+y^{3}=3 a x y(a>0)$( 提示 : 令 $y=x t$ )}
\textbf{解}\quad
令$y=xt$可得:$x=\dfrac{3at}{1+t^3},y=\dfrac{3at^2}{1+t^3}$
$$
	\begin{aligned}
		S & =\dfrac{1}{2}\oint\limits_Lx\mathrm{d}y-y\mathrm{d}x             \\
		  & =\dfrac{9a^2}{2}\int_0^{\infty}\dfrac{t^2}{(1+t^3)^2}\mathrm{d}t \\
		  & =\dfrac{3a^2}{2}
	\end{aligned}
$$

\subsection{抛物线 $(x+y)^{2}=a x(a>0)$ 与 $x$ 轴所围成的图形}
\textbf{解}\quad
令$x+y=t,$则$x=\dfrac{t^2}{a},y=t-\dfrac{t^2}{a}$.
$$
	\begin{aligned}
		S & =\dfrac{1}{2}\oint\limits_Lx\mathrm{d}y-y\mathrm{d}x \\
		  & =\dfrac{1}{2}\int_0^a\dfrac{t^2}{a}\mathrm{d}t        \\
		  & =\dfrac{a^2}{6}
	\end{aligned}
$$

\section{设封闭曲线 $L$ 有参数方程 $x=\varphi(t), y=\psi(t), t \in[\alpha, \beta]$, 参数增加时指示 $L$ 的正向. 证明 $: L$ 围成的面积 $A=\dfrac{1}{2} \int\limits_{a}^{\beta}\left|\begin{array}{ll}\varphi(t) & \psi(t) \\ \varphi^{\prime}(t) & \psi^{\prime}(t)\end{array}\right| \mathrm{d} t $}
\begin{proof}
	格林公式的简单代换:
	$$
		\begin{aligned}
			\dfrac{1}{2} \int\limits_{a}^{\beta}\left|\begin{array}{ll}\varphi(t) & \psi(t) \\ \varphi^{\prime}(t) & \psi^{\prime}(t)\end{array}\right| \mathrm{d} t & =\dfrac{1}{2}\oint\limits_Lx\mathrm{d}y-y\mathrm{d}x \\
			                                                                                         & =S
		\end{aligned}
	$$
\end{proof}



\section{计算曲线积分 $\oint\limits_{L} \dfrac{\mathrm{e}^{x}}{x^{2}+y^{2}}[(x \sin y-y \cos y) \mathrm{d} x+(x \cos y+y \sin y) \mathrm{d} y]$, 其中 $L$ 是包含原点在其内部的分段光滑的闭曲线}
\textbf{解}\quad
原点处无定义,从而令$L1:x^2+y^2=r^2$,方向向外.
$$
	P=\dfrac{\mathrm{e}^x(x\sin y - y\cos y)}{x^2+y^2},Q=\dfrac{\mathrm{e}^x(x\cos y + y\sin y)}{x^2+y^2}
$$
且有
$$
	\begin{aligned}
		\dfrac{\partial Q}{\partial x} & =\mathrm{e}^x\dfrac{[(x^2+y^2)x+y^2-x^2]\cos y+(x^2+y^2-2x)y\sin y}{(x^2+y^2)^2} \\
		\dfrac{\partial P}{\partial y} & =\mathrm{e}^x\dfrac{[(x^2+y^2)x+y^2-x^2]\cos y+(x^2+y^2-2x)y\sin y}{(x^2+y^2)^2}
	\end{aligned}
$$
从而$\nabla \times (P,Q)=0$,从而有
$$
	\begin{aligned}
		\oint\limits_{L} \dfrac{\mathrm{e}^{x}}{x^{2}+y^{2}}[(x \sin y-y \cos y) \mathrm{d} x+(x \cos y+y \sin y) \mathrm{d} y] & =\oint\limits_{L1}P\mathrm{d}x+Q\mathrm{d}y                            \\
		                                                                                                                        & =\dfrac{1}{r^2}\iint\limits_S2\mathrm{e}^x\cos y\mathrm{d}x\mathrm{d}y
	\end{aligned}
$$
令$r\rightarrow 0$可得
$$
	\begin{aligned}
		\lim_{r\rightarrow 0}\dfrac{1}{r^2}\iint\limits_S2\mathrm{e}^x\cos y\mathrm{d}x\mathrm{d}y & =\lim_{r\rightarrow 0}\dfrac{\pi r^2 2\mathrm{e}^{\xi_1}\cos \xi_2}{r^2}
	\end{aligned}
$$
由于$\xi_1,\xi_2<r$,从而
$$
	\lim_{r\rightarrow 0}\dfrac{1}{r^2}\iint\limits_S2\mathrm{e}^x\cos y\mathrm{d}x\mathrm{d}y=2\pi
$$

\section{求下列全微分的原函数}
\subsection{$\left(x^{2}-2 x y+y^{2}\right) \mathrm{d} x-\left(x^{2}-2 x y-y^{2}\right) \mathrm{d} y$}
\textbf{解}\quad
$P=(x^2-2xy+y^2),Q=-(x^2-2xy-y^2)$.有
$$
	\begin{aligned}
		\dfrac{\partial P}{\partial y} & =-2x+2y \\
		\dfrac{\partial Q}{\partial x} & =-2x+2y
	\end{aligned}
$$
从而$\nabla \times (P,Q)=0$,积分与路径无关.所以取$(x_0,y_0)=(0,0)$,则全体原函数可表示为
$$
	\begin{aligned}
		F(x,y) & =\int_{(0,0)}^{(x,y)}\left(x^{2}-2 x y+y^{2}\right) \mathrm{d} x-\left(x^{2}-2 x y-y^{2}\right) \mathrm{d} y
	\end{aligned}
$$
以折线$(0,0)\rightarrow(0,y)\rightarrow(x,y)$为路径进行积分:
$$
	\begin{aligned}
		F(x,y) & =\int_0^{y}y^2\mathrm{d}y+\int_0^x(x^2-2xy+y^2)\mathrm{d}x \\
		       & =\dfrac{y^3}{3}+\dfrac{x^3}{3}-x^2y+xy^2+C
	\end{aligned}
$$
\subsection{$x f\left(\sqrt{x^{2}+y^{2}}\right) \mathrm{d} x+y f\left(\sqrt{x^{2}+y^{2}}\right) \mathrm{d} y$}
\textbf{解}\quad
$P=xf\left(\sqrt{x^2+y^2}\right),Q=yf\left(\sqrt{x^2+y^2}\right)$,且有
$$
	\begin{aligned}
		\dfrac{\partial P}{\partial y} & =\dfrac{xyf'}{\sqrt{x^2+y^2}} \\
		\dfrac{\partial Q}{\partial x} & =\dfrac{xyf'}{\sqrt{x^2+y^2}}
	\end{aligned}
$$
从而$\nabla \times (P,Q)=0$,所以与积分路径无关.

令$f(x,y)=f\left(\sqrt{x^2+y^2}\right)(x\mathrm{d}x+y\mathrm{d}y)=f\left(\sqrt{x^2+y^2}\right)\dfrac{1}{2}\mathrm{d}(x^2+y^2)=uf(u)\mathrm{d}u$.

从而$\sqrt{x^2+y^2}f\left(\sqrt{x^2+y^2}\right)$的原函数即为所求.

\section{先证明下列曲线积分与路径无关,再计算其积分值}
\subsection{$\int_{(2,1)}^{(1.2)} \dfrac{y \mathrm{~d} x-x \mathrm{~d} y}{x^{2}}$, 沿在右半平面的路线}
\textbf{解}\quad
$$
	\begin{aligned}
		\dfrac{\partial P}{\partial y} & =\dfrac{1}{x^2} \\
		\dfrac{\partial Q}{\partial x} & =\dfrac{1}{x^2}
	\end{aligned}
$$
从而$\nabla \times (P,Q)=0$,积分与路径无关.沿折线积分:
$$
	\begin{aligned}
		I & =\int_2^1\dfrac{1}{x^2}\mathrm{d}x-\int_1^2\mathrm{d}y \\
		  & =-\dfrac{3}{2}
	\end{aligned}
$$

\subsection{$\int_{(2, 1)}^{(1,2)} \varphi(x) \mathrm{d} x+\psi(y) \mathrm{d} y, \varphi(x), \psi(y)$ 为连续函数}
\textbf{解}\quad
$$
	\begin{aligned}
		\dfrac{\partial P}{\partial y} & =0 \\
		\dfrac{\partial Q}{\partial x} & =0
	\end{aligned}
$$
从而$\nabla \times (P,Q)=0$,积分与路径无关.沿折线积分:
$$
	\begin{aligned}
		I & =\int_2^1\varphi(x)\mathrm{d}x\int_1^2\psi(y)\mathrm{d}y
	\end{aligned}
$$

\section{已知平面区域 $D=\{(x, y) \mid 0 \leqslant x \leqslant 1,0 \leqslant y \leqslant 1\}, L$ 为 $D$ 的正向边界, $f(x)$ 为$[0,1]$ 上的连续函数,证明}
\subsection{$\oint\limits_{L} x \mathrm{e}^{f(y)} \mathrm{d} y-y \mathrm{e}^{-f(x)} \mathrm{d} x=\oint\limits_{L} x \mathrm{e}^{-f(y)} \mathrm{d} y-y \mathrm{e}^{f(x)} \mathrm{d} x$}
\begin{proof}
	$$
		\begin{aligned}
			\oint\limits_{L} x \mathrm{e}^{f(y)} \mathrm{d} y-y \mathrm{e}^{-f(x)} \mathrm{d} x & =\iint\limits_{D}\left(\mathrm{e}^{f(y)}+\mathrm{e}^{-f(x)}\right) \mathrm{d} x \mathrm{~d} y \\
			\oint\limits_{L} x \mathrm{e}^{-f(y)} \mathrm{d} y-y \mathrm{e}^{f(x)} \mathrm{d} x & =\iint\limits_{D}\left(\mathrm{e}^{-f(y)}+\mathrm{e}^{f(x)}\right) \mathrm{d} x \mathrm{~d} y
		\end{aligned}
	$$
	因为区域 $D$ 关于 $y=x$ 对称,所以由轮换对称性可知
	$$
		\iint\limits_{D}\left(\mathrm{e}^{f(y)}+\mathrm{e}^{-f(x)}\right) \mathrm{d} x \mathrm{~d} y=\iint\limits_{D}\left(\mathrm{e}^{-f(y)}+\mathrm{e}^{f(x)}\right) \mathrm{d} x \mathrm{~d} y
	$$
	$$
		\oint\limits_{L} x \mathrm{e}^{f(y)} \mathrm{d} y-y \mathrm{e}^{-f(x)} \mathrm{d} x=\oint\limits_{L} x \mathrm{e}^{-f(y)} \mathrm{d} y-y \mathrm{e}^{f(x)} \mathrm{d} x
	$$
\end{proof}

\subsection{$\oint\limits_{L} x \mathrm{e}^{f(y)} \mathrm{d} y-y \mathrm{e}^{-f(x)} \mathrm{d} x \geqslant 2$}
\begin{proof}
    $$
    \begin{aligned}
            \oint\limits_{L} x \mathrm{e}^{f(y)} \mathrm{d} y-y \mathrm{e}^{-f(x)} \mathrm{d} x&=\iint\limits_{D}\left(\mathrm{e}^{f(y)}+\mathrm{e}^{-f(x)}\right) \mathrm{d} x \mathrm{~d} y\\
            &\geqslant 2\iint\limits_D\mathrm{d}x\mathrm{d}y\\
            &=2
    \end{aligned}
    $$
\end{proof}

\section{设函数 $f(x), g(x)$ 具有 $2$ 阶连续导数,并且积分$$\oint\limits_{L}\left(y^{2} f(x)+2 y \mathrm{e}^{x}+2 y g(x)\right) \mathrm{d} x+2(y g(x)+f(x)) \mathrm{d} y=0$$对平面上任一条封闭曲线 $L$ 成立. 求 $f(x), g(x) .$}
\textbf{解}\quad
$P=y^{2} f(x)+2 y \mathrm{e}^{x}+2 y g(x),Q=2(y g(x)+f(x))$
由于积分与路径无关,从而
$$
yf(x)+g(x)+\mathrm{e}^x=yg'(x)+f'(x)
$$
令$y=2,1$,有
\begin{equation}
    2f(x)+g(x)+\mathrm{e}^x=2g'(x)+f'(x)
\end{equation}
\begin{equation}
    f(x)+g(x)+\mathrm{e}^x=g'(x)+f'(x)
\end{equation}
由$(1)-(2)$可得:
$f(x)=g'(x),g(x)+\mathrm{e}^x=f'(x)$.从而$g''-g=\mathrm{e}^x$.

解得$g(x)=\dfrac{1}{2}x\mathrm{e}^x+c\mathrm{e}^x+d\mathrm{e}^{-x},f(x)=\dfrac{1}{2}\mathrm{e}^x(1+x)+c\mathrm{e}^x-d\mathrm{e}^x$.

\end{document}